单位圆的最多覆盖据点 （重点需要学习！）

我自己的代码，提交错误，需要修改

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#include <math.h>

#include <ctype.h>

#include <iostream>

#include <string>

#include <algorithm>

#define eps 1e-8



using namespace std;



struct point

{

	double x;

	double y;

	int w;

	point(){}

	point (double tx, double ty)

	{

	    x=tx; y=ty;

	}

}a[300+10];



double Dist(point a, point b )

{

	return sqrt( (b.x-a.x)*(b.x-a.x) + (b.y-a.y)*(b.y-a.y) );

}



point find_start(point p1, point p2)

{

/*	point p, mid, start;

	double d, aa;

	p.x=p2.x-p1.x;

	p.y=p2.y-p1.y;



	mid.x = (p1.x +p2.x)/2;

	mid.y = (p1.y +p2.y)/2;



	d = sqrt( 1-Dist(p1, mid) );//公共弦长的一半长



	if( fabs(p.y) < eps )

	{

		start.x = mid.x;

		start.y = mid.y + d;

	}

	else

	{

		aa = atan(-p.x/p.y);

		start.x = mid.x + d*cos(aa);

		start.y = mid.y + d*sin(aa);

	}

	return start; */

	point mid=point((p1.x+p2.x)/2,(p1.y+p2.y)/2);



    double angle=atan2(p1.x-p2.x,p2.y-p1.y);



    double d=sqrt(1-Dist(p1,mid)*Dist(p1,mid));



    return point(mid.x+d*cos(angle), mid.y+d*sin(angle));

}





int main()

{

    int n;

    int i, j, k;



	int ans, ans0;

	point centre;

	double tmp;



  	while(scanf("%d", &n)!=EOF)

	{

		if(n==0) break;

		for(i=0; i<n; i++)

		{

			scanf("%lf %lf %d", &a[i].x, &a[i].y, &a[i].w );

		}

		ans=0; //结果至少为1

		for(i=0; i<n; i++)

		{

			for(j=i+1; j<n; j++)

			{

				if( Dist(a[i], a[j]) > 2.0 )

					continue;

				ans0=0; //

				centre = find_start(a[i], a[j]);



				for(k=0; k<n; k++)

				{

					tmp = Dist(centre, a[k]);

					if(tmp<1.00000001)

						ans0+=a[k].w;

				}

				ans=max(ans, ans0);

			}

		}

		printf("%d\n", ans );

	}

	return 0;

}

 上述方法理论上的处理数据量不大

还有另外一种方法：可以参考如下博客

http://blog.csdn.net/acm_cxlove/article/details/7894310

 

hihocoder #1064 解题博客：地址：http://blog.csdn.net/u014076176/article/details/39825957

#include<stdio.h>

#include<iostream>

#include<string.h>

#include<math.h>

#include<algorithm>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<string>

#define ll long long

#define db double

#define PB push_back

#define lson k<<1

#define rson k<<1|1

using namespace std;



const int N = 2005;

const db PI = acos(-1.0);

const db eps = 1e-8;



int sgn(db t)

{

    return t<-eps?-1:t>eps;

}



struct Point

{

    db x,y;

    int w;

    Point (db _x=0,db _y=0):x(_x),y(_y) {}

    void input()

    {

        scanf("%lf%lf%d",&x,&y,&w);

    }

    db len2()

    {

        return x*x+y*y;

    }

    db len()

    {

        return sqrt(len2());

    }

    Point operator - (const Point &t) const

    {

        return Point(x-t.x,y-t.y);

    }

    bool operator == (const Point &t) const

    {

        return sgn(x-t.x)==0&&sgn(y-t.y)==0;

    }

    db operator * (const Point &t) const

    {

        return x*t.y-t.x*y;

    }

} p[N];



struct node

{

    db thta;

    int w;

    bool operator < (const node &t) const

    {

        if(thta==t.thta) return w>t.w;

        return thta<t.thta;

    }

} jd[N*4];



int ln;

void add(db thta,int w)

{

    jd[ln].thta=thta,jd[ln++].w=w;

    jd[ln].thta=thta+PI*2,jd[ln++].w=w;

}



int main()

{

    int n;

    scanf("%d",&n);

    for(int i=0; i<n; i++) p[i].input();

    Point px;

    int ans=0;

    for(int i=0; i<n; i++)

    {

        px=Point(p[i].x+1.0,p[i].y);

        ln=0;

        int nw=p[i].w;

        for(int j=0; j<n; j++)

        {

            if(i!=j)

            {

                db d=(p[i]-p[j]).len();

                if(d>2.0) continue;

                db thta=atan2((p[j]-p[i]).y,(p[j]-p[i]).x);

                db th=acos(d/2.0);

                add(thta-th,p[j].w),add(thta+th,-p[j].w);

            }

        }

        sort(jd,jd+ln);

        int mm=nw;

        for(int j=0; j<ln; j++)

            nw+=jd[j].w,mm=max(mm,nw);

        ans=max(ans,mm);

    }

    printf("%d\n",ans);

    return 0;

}